Dividing $R$ at $\frac{2}{3}, \frac{4}{3}$ and $2$, analyses $4$ cases.
When $x \leq \frac{2}{3}$, the inequality becomes
$2-3 x+4-3 x+6-3 x \geq 12$.
implying $-9 x \geq 0 \Rightarrow x \leq 0$.
when $x \ge 2$ the ineqality becomes
$3 x-2+3 x-4+3 x-6 \geq 12$
Implying $9 x \geq 24 \Rightarrow x \geq 8 / 3$
The inequality in invalid in the other two sections.
$\therefore $ either $x \leq 0$ or $x \geq 8 / 3$